[LeetCode] Evaluate Reverse Polish Notation 计算逆波兰表达式

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are

+

,

-

,

*

,

/

. Each operand may be an integer or another expression.

Some examples:

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

逆波兰表达式就是把操作数放前面，把操作符后置的一种写法，我们通过观察可以发现，第一个出现的运算符，其前面必有两个数字，当这个运算符和之前两个数字完成运算后从原数组中删去，把得到一个新的数字插入到原来的位置，继续做相同运算，直至整个数组变为一个数字。于是按这种思路写了代码如下：

/**
* Time Limit Exceeded
*/
class Solution {
public:
int evalRPN(vector<string> &tokens) {
if (tokens.size() == 1) return atoi(tokens[0].c_str());
int n = tokens.size();
int cur = 0, res = 0;
while (tokens.size() != 1) {
n = tokens.size();
cur = 0;
while (tokens[cur] != "+" && tokens[cur] != "-" && tokens[cur] != "*" && tokens[cur] != "/") ++cur;
int a = atoi(tokens[cur - 2].c_str());
int b = atoi(tokens[cur - 1].c_str());
if (tokens[cur] == "+") res = a + b;
if (tokens[cur] == "-") res = a - b;
if (tokens[cur] == "*") res = a * b;
if (tokens[cur] == "/") res = a / b;
tokens.insert(tokens.begin() + cur + 1, to_string(res));
for (int i = cur; i >= cur - 2; --i) {
tokens.erase(tokens.begin() + i);
}
}
return res;
}
};

但是拿到OJ上测试，发现会有Time Limit Exceeded的错误，无奈只好上网搜答案，发现大家都是用栈做的。仔细想想，这道题果然应该是栈的完美应用啊，从前往后遍历数组，遇到数字则压入栈中，遇到符号，则把栈顶的两个数字拿出来运算，把结果再压入栈中，直到遍历完整个数组，栈顶数字即为最终答案。代码如下:

class Solution {
public:
int evalRPN(vector<string> &tokens) {
if (tokens.size() == 1) return atoi(tokens[0].c_str());
stack<int> s;
for (int i = 0; i < tokens.size(); ++i) {
if (tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/")
　　　　　　　{
s.push(atoi(tokens[i].c_str()));
} else {
int m = s.top();
s.pop();
int n = s.top();
s.pop();
if (tokens[i] == "+") s.push(n + m);
if (tokens[i] == "-") s.push(n - m);
if (tokens[i] == "*") s.push(n * m);
if (tokens[i] == "/") s.push(n / m);
}
}
return s.top();
}
};

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